JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Current Electricity

When two resistance

R_{1}

and

R_{2}

connected in series and introduced into the left gap of a meter bridge and a resistance of

10 Ω

is introduced into the right gap, a null point is found at 60 cm from left side. When

R_{1}

and

R_{2}

are connected in parallel and introduced into the left gap, a resistance of

3 Ω

is introduced into the right-gap to get null point at 40 cm from left end. The product of

R_{1} R_{2}

is ______

Ω^{2}

Answer: 30

Solution

In a metre bridge the balance condition sets the left-gap resistance against the right. With the two resistors in series,

\frac{R _{1} + R _{2}}{10} = \frac{60}{40}

, so

R_{1} + R_{2} = 15 Ω

. With them in parallel against

3 Ω

and the null at

40

cm,

\frac{R _{1} R _{2} / ( R _{1} + R _{2} )}{3} = \frac{40}{60}

, which gives

\frac{R _{1} R _{2}}{15 \times 3} = \frac{2}{3}

. Solving,

R_{1} R_{2} = 30 Ω^{2}

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer